Now its very strenuous

Algebra Level 4

Three positive real numbers $a$ , $b$ , and $c$ satisfy the following equation. $\large a+b+c=abc$ Let $\large N = 3+\dfrac{b}{a^2}+\dfrac{c}{b^2}+\dfrac{a}{c^2}-\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2$ If the minimum value of $N$ can be expressed as $\sqrt{x}$ , where $x$ is square free positive integer. Find $x$ .

The answer is 3.

1 solution

Atul Shivam
Apr 24, 2016

Assume that all the digits $a,b,c$ are the same, ie. $a=b=c=k$ ,where $k$ arbitrary real number.

$k+k+k=k^3$ $3k=k^3$ $k=\sqrt {3}$

Now $N=3+\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2$

replacing variables with $`k`$ gives $N=3+\frac{k}{k²}+\frac{k}{k^2}+\frac{k}{k^2}-(\frac{1}{k}+\frac{1}{k}+\frac{1}{k})^2$

Now,putting $k=`\frac{1}{\sqrt{3}}`$ we get $N=3+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2=3+\sqrt{3}-3=\sqrt{3}$ $\sqrt{x}=\sqrt{3}$ $x=\boxed{3}$

Moderator note:

This is not a proper solution.

And what if all of the digits are not the same?

Calvin Lin Staff - 5 years, 1 month ago

But I see nothing is mentioned in the question whether $a,b,c$ are same or not, that is why I approached this way.

Thanks!!!

Atul Shivam - 5 years, 1 month ago

Now its very strenuous

The answer is 3.

1 solution

Moderator note:

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